array chip wrote:
Looks like the log file is not appropriately attached. Here it is again.
Thanks
for any suggestions.
John
Well, SAS isn't finding the file test in the SAS library
.\Desktop, presumably because it has the wrong extension.
The obvious question is whether you can read
Cristian Montes wrote:
Hello,
I am trying to plot confidence bands on the mean and prediction bands for the
following
nonlinear regression, using maximum likelihood via optim. A toy example with
data and
code of what I am trying to accomplish is:
VOL-c(0.01591475, 1.19147935
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Peter Dalgaard
Sent: Monday, August 02, 2010 11:30 PM
To: array chip
Cc: r-help@r-project.org
Subject: Re: [R] read SAS dataset using read.ssd()
array chip wrote:
Looks
I made some matlab codes...
Is there any method to perform matlab codes in R program??
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
__
Then just use order() for instance,
mydata - read.table(textConnection(
20071130 information info
20071031 information info))
closeAllConnections()
mydata[order(strptime(mydata$V1, format = %Y%m%d)), ]
HTH,
Josh
On Mon, Aug 2, 2010 at 7:07 PM, Leigh E. Lommen
sort_df in package reshape?
On Mon, Aug 2, 2010 at 9:07 PM, Leigh E. Lommen
leigh.lom...@courtesycorporation.com wrote:
a - c( 20071031,20071130, 20071231)
sort(a)
Or if you want convert to date:
sort(strptime(a, '%Y%m%d'))
Okay, but what if you have information in the other columns?
I sometimes had trouble importing data that was exported from Excel.
Exporting through OpenOffice sometimes resolved the issue.
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Sent from the R help mailing list archive at
Addi Wei addi...@gmail.com writes:
Hello,
I am having some trouble using a model I created from plsr (of train) to
analyze each invididual R^2 of the 10 components against the test data. For
example:
mice1 - plsr(response ~factors, ncomp=10 data=MiceTrain)
R2(mice1)##this
I assume you will need
.Fortran(cov1, .., PACKAGE=RcmdrPlugin.push)
if the fortran code is in your package.
Best,
Uwe Ligges
On 03.08.2010 05:54, Erin Hodgess wrote:
Dear R People:
Hello!
I'm putting together another RcmdrPlugin package and need to add a
FORTRAN subroutine to speed
Dear r-helpers,
I want to randomly sample n points from regions of a raster layers,
the cells denoted as NA is not
included in this sampling process. And, I want to got the longitude
and latitude of the sampled points.
I checked the manual of raster package, I found several functions is
relative
On Tue, 3 Aug 2010 10:00:08 +1000
Glen Barnett glnbr...@gmail.com wrote:
This might help some:
RSiteSearch(winsorize)
Or
require(sos)
findFn(winsorize)
Liviu
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
On Aug 3, 2010, at 12:05 AM, Daniel Malter wrote:
Hi, I have a good grasp of grep() and gsub() for finding and
extracting
character strings. However, I cannot figure out how to use a search
term
that is stored in a variable when the search string is more complex.
#Say I have a string,
Hi All,
I'm trying to run the following script in R, but I'm getting a warning saying:
Warning message:
In if (z 0) { :
the condition has length 1 and only the first element will be used
As you can see, I'm sending a vector x to the function f without any
problem. The function f calculates
Hi
r-help-boun...@r-project.org napsal dne 03.08.2010 13:03:33:
Hi All,
I'm trying to run the following script in R, but I'm getting a warning
saying:
Warning message:
In if (z 0) { :
the condition has length 1 and only the first element will be used
As you can see, I'm sending
Hello, Pablo,
if() doesn't accept a vector as its argument the way you may want it to;
take a look at ifelse().
Regards -- Gerrit
On Tue, 3 Aug 2010, Pablo Cerdeira wrote:
Hi All,
I'm trying to run the following script in R, but I'm getting a warning saying:
Warning message:
In if (z
Ok thanks, but I have found the solution: ifelse.
Here goes a working version of the code:
mod = function(x) {
ifelse(x 0,x*(-1),x)
}
f = function(x) {
f - mod(x)/x
}
x - seq(-1,1,0.01)
x
y - f(x)
y
plot(f,xlim = c(-1,1))
remove(x,y,f,mod)
best regards
On Tue, Aug 3, 2010 at 8:03 AM,
Many thanks to Ben and Peter for their help. Even if I am not really
happy with it, I used the environment solution and got the results I
expected.
Issue #2 is stickier. I think I must say that the idelology of mle() is
that the user passes a likelihood function. If the likelihood function
On Aug 3, 2010, at 7:03 AM, Pablo Cerdeira wrote:
Hi All,
I'm trying to run the following script in R, but I'm getting a
warning saying:
Warning message:
In if (z 0) { :
the condition has length 1 and only the first element will be used
ifelse is the proper function rather than
Translate the matlab code to the appropriate R expressions.
On Mon, Aug 2, 2010 at 9:57 PM, leepama butch...@hanmail.net wrote:
I made some matlab codes...
Is there any method to perform matlab codes in R program??
--
View this message in context:
Look at the help page for 'if':
A length-one logical vector that is not NA. Conditions of length
greater than one are accepted with a warning, but only the first
element is used. Other types are coerced to logical if possible,
ignoring any class.
If you want the positive value, why aren't you
On Mon, Aug 2, 2010 at 9:57 PM, leepama butch...@hanmail.net wrote:
I made some matlab codes...
Is there any method to perform matlab codes in R program??
There is an Octave/R lexicon here:
http://cran.r-project.org/doc/contrib/R-and-octave.txt
There are also ROctave and R.matlab
I would like to be able to grab x and y columns out of a dataframe and
then grab all of the columns that are not equal to x or y. I am sure
that I am missing something easy.
ftbr_UTM_downstream - (structure(list(site =
c(Jennie_Creek_Main_Stem, Wolf_Pit_Creek_Main_Stem,
#this does it sorry for clogging everyones email boxes
charmatch(c(x, y), colnames(ftbr_UTM_downstream))
On Tue, Aug 3, 2010 at 6:41 AM, stephen sefick ssef...@gmail.com wrote:
I would like to be able to grab x and y columns out of a dataframe and
then grab all of the columns that are not
On Aug 2, 2010, at 9:57 PM, leepama wrote:
I made some matlab codes...
Is there any method to perform matlab codes in R program??
Octave/Matlab equivalencies:
http://cran.r-project.org/doc/contrib/R-and-octave.txt
http://rosettacode.org/wiki/Main_Page
David Winsemius, MD
West
Hi all
My data table (g) contains a continues data column (plant.height) and other
columns (columns 8 to 57),
each with number of levels of different factors. ANOVA test was done and
the p-values were extracted
as follos:
a - function(x) anova(lm(plant.height ~ x))$Pr(F)[1]
r-
On Aug 3, 2010, at 7:41 AM, stephen sefick wrote:
I would like to be able to grab x and y columns out of a dataframe and
then grab all of the columns that are not equal to x or y. I am sure
that I am missing something easy.
ftbr_UTM_downstream - (structure(list(site =
Hi R-Help.
I am working on a data set with a 3-level nested structure. I have
individuals nested in households and multiple observations on each
individual. I assume that the individuals inside a given household are
correlated and that the individuals are correlated with themselves
over time. The
Thank you!
Now, here is the code for the FORTRAN subroutine:
subroutine cov1(n1,na,x,z,yy,ac)
integer n1,na
real x(n1),ac(na+1),z,yy
do 5 j=1,(na+1)
do 10 i=1,(n1-j)
if(j.eq.1)ac(j)=yy
if(j.ne.1)ac(j)=ac(j)+(x(i)-z)*(x(i+j-1)-z)
10
As a matter of fact, I would say both Bert and I encounter designed
experiments a lot more than observational studies, yet we speak from
experience that those things that Bert mentioned happen on a daily
basis. When you talk to experimenters, ask your questions carefully and
you'll see these
Hi
r-help-boun...@r-project.org napsal dne 03.08.2010 13:17:16:
Ok thanks, but I have found the solution: ifelse.
Here goes a working version of the code:
mod = function(x) {
ifelse(x 0,x*(-1),x)
}
But before you reinvent wheel try to look if there is not available
function already
Dear all,
I have one dependent variable y and two independent variables x1 and x2
which I would like to use to explain y. x1 and x2 are design factors in an
experiment and are not correlated with each other. For example assume that:
x1 - rbind(1,1,1,2,2,2,3,3,3)
x2 - rbind(1,2,3,1,2,3,1,2,3)
Are x1 and x2 are factors (dummy variables)? cor does not make sense
in this case.
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail.com
On Aug 3, 2010, at 9:10 AM, Michael Haenlein wrote:
Dear all,
I have one dependent variable y and
Thanks for your comment!
Actually, they are continuous variables which have a very low correlation --
I just wanted to make the whole story easier for explanation.
My general question is: Does R offer an alternative to lm for situations
where there is substantial collinearity between the
I think you are attributing to collinearity a problem that is due to
your small sample size. You are predicting 9 points with 3 predictor
terms, and incorrectly concluding that there is some inconsistency
because you get an R^2 that is above some number you deem surprising.
(I got values
I'm sorry -- I think I chose a bad example. Let me start over again:
I want to estimate a moderated regression model of the following form:
y = a*x1 + b*x2 + c*x1*x2 + e
Based on my understanding, including an interaction term (x1*x2) into the
regression in addition to x1 and x2 leads to
Hello,
I use the following code to generate this plot:
http://imgur.com/GsWvY.jpg. How do I remove the labels to the left of
the middle axis? Or is there a simpler solution that gets me the same
plot. I basically just want to highlight 25.67 on the leftmost axis, and
have all other labels
David,
Thanks. It turns out that, once you've set up your locales properly, it's
almost impossible to create an example for the problem.
I'm working with scientific text which contains a fair amount of symbols:
degrees, plus-or-minus, etc. When I read the text in, I specified UTF-8. My
On Aug 3, 2010, at 9:51 AM, haenl...@gmail.com wrote:
I'm sorry -- I think I chose a bad example. Let me start over again:
I want to estimate a moderated regression model of the following form:
y = a*x1 + b*x2 + c*x1*x2 + e
Based on my understanding, including an interaction term (x1*x2)
My usual strategy of dealing with multicollinearity is to drop the
offending variable or transform one them. I would also check vif
functions in car and Design.
I think you are looking for lm.ridge in MASS package.
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North
Hi Everyone,
I found the doe.base package and the FrF2 package to do nice experimental
planning and I'm very happy about this tool I was looking for such a long
time.
But I still try to find out how to add center points to a full factorial
design. The FrF2-package has a center point option but
HI, I am really messing up to make a symmetrical matrix using upper.tri()
lower.tri() function. Here is my code:
set.seed(1)
mat = matrix(rnorm(25), 5, 5)
mat
[,1] [,2] [,3] [,4] [,5]
[1,] -0.6264538 -0.8204684 1.5117812 -0.04493361 0.91897737
[2,]
I would suggest to you the following:
1) Run the same thing, but with a loop instead of apply
2) add the to loop a printing that shows you on what cycle of the loop the
function breaks
3) see if that vector has any Inf or NA values (although in general I think
you are using a numeric instead of a
Thanks very much -- it seems that Ridge Regression can do what I'm looking
for!
Best,
Michael
-Original Message-
From: Nikhil Kaza [mailto:nikhil.l...@gmail.com]
Sent: Tuesday, August 03, 2010 16:21
To: haenl...@gmail.com
Cc: r-help@r-project.org (r-help@R-project.org)
Subject: Re: [R]
If the collinearity you're seeing arose from the addition of a product
(interaction) term, I do not think penalization is the best answer.
What is the goal of your analysis? If it's prediction, then I wouldn't
worry about this type of collinearity. If you're interested in
inference, I'd try some
Dear all,
I am using GNU screen to run multiple R sessions from one working
directory in order to split task, however I noticed that dataset is not
synchronized e.g. if I have two sessions R1 and R2, and I remove an
object from R1, R2 doesn't change as expected or change at random.
I have tried
Dear list,
is there a way to open a .zip folder so that one can extract and
modify files inside and then save teh .zip folder again?
thanks!
Jose
--
Dr. Jose I. de las Heras Email: j.delashe...@ed.ac.uk
The Wellcome Trust Centre for Cell BiologyPhone: +44 (0)131
On Aug 3, 2010, at 2:41 PM, Liaw, Andy wrote:
As a matter of fact, I would say both Bert and I encounter designed
experiments a lot more than observational studies, yet we speak from
experience that those things that Bert mentioned happen on a daily
basis. When you talk to experimenters,
The apply function coerces the factor results to a character array
apply(g,2,class) # gives character
The kruskal.test function doesn't take character vector as the group
argument.
kruskal.test(as.character(plant.height) ~ as.character(g[,8])) #doesn't work
kruskal.test(plant.height ~
At 19:31 02/08/2010, wwreith wrote:
I am testing normality on the studetized residuals that are
generated after performing ANOVA and yes I used Levene's test to see
if the variances can be assumed equal. They infact are not, but I
have found a formula for determining whether the p-value for
Found it out by myself: If you try FrF2(4,2, ncenter=X) you get a full
factorial design.
Sorry for that. Should have tried before.
But maybe this will help somebody else, too.
Greetings, Tobias
--
View this message in context:
On Aug 3, 2010, at 1:47 PM, imrib wrote:
Hi all
My data table (g) contains a continues data column (plant.height) and other
columns (columns 8 to 57),
each with number of levels of different factors. ANOVA test was done and
the p-values were extracted
as follos:
a - function(x)
On Aug 3, 2010, at 7:36 AM, Ron Michael wrote:
HI, I am really messing up to make a symmetrical matrix using
upper.tri() lower.tri() function. Here is my code:
set.seed(1)
mat = matrix(rnorm(25), 5, 5)
mat
[,1] [,2] [,3][,4][,5]
[1,] -0.6264538
I have an array called stocks which contains numeric dates, ticker
symbols,prices, etc.
stocks[1:3,]
DATETICKER PERMNO EXCHCD TSYMBOL TRDSTAT SHROUT PRC
RET
1 19950131 EWST 10001 3 EWST A
2224 -7.75000 -0.031250
2 19950228 EWST 10001
Dear list,
I have a list of matrices :
i1 - matrix(1:10, nrow = 2, ncol = 5)
i2 - matrix(11:20, nrow = 2, ncol = 5)
j - list(i1 = i1, i2 = i2)
I would like to attribute names to each dimension, for each matrix,
as follows :
$i1
B1 B2 B3 B4 B5
A1 1 3 5 7 9
A2 2 4 6 8 10
$i2
B1 B2 B3 B4 B5
A1
thanks everyone. I will try them.
Best Wishes
Nan
- Original Message
From: Steven McKinney smckin...@bccrc.ca
To: Hey Sky heyskywal...@yahoo.com; r-help@r-project.org
r-help@r-project.org
Sent: Mon, August 2, 2010 10:56:32 PM
Subject: RE: [R] how to do a IF ELSE in a matrix
Hi folks,
CPU - AMD X4 955
Onboard RAM - 8G
I'm prepared to learn R and am going to install R package on 64 bit Windows
running as VM (guest) on Oracle VituralBox. Please advise
1) I have 64 bit Win7 and Win Server 2008 R2 running as VM on VirtualBox.
Which
version of Windows will be more
Hi Carlos,
Look at this:
i1 - matrix(1:10, nrow = 2, ncol = 5)
i2 - matrix(11:20, nrow = 2, ncol = 5)
j - list(i1 = i1, i2 = i2)
my.names - list(c(A1, A2), c(B1, B2, B3, B4, B5))
j.new - lapply(j, function (x) {dimnames(x) - my.names; return(x)})
j.new
the issue was just that you have to save
Hi, all
There are 62 samples in my data and I tested 3 times for each one, then I
want to use ICC(intraclass correlation) from irr package to test the
consistency among the tests.
*combatexpdata_p[1:62] is the first text results and combatexpdata_p[63:124]
* is the second one and
try using Matrix package instead
mat - Matrix(rnorm(25),5,5)
forceSymmetric(mat)
The reason your method does not work is because matrix is effectively
a vector and the indices increase along rows within a column.
Nikhil
On Aug 3, 2010, at 7:36 AM, Ron Michael wrote:
HI, I am really
Hi Chen,
From the documentation (see ?icc) $value: the intraclass correlation
coefficient.
As an example:
set.seed(1)
result - icc(matrix(rnorm(20), ncol = 2))
result$value
[1] -0.3148786
HTH,
Josh
On Tue, Aug 3, 2010 at 8:50 AM, chen chao chench...@gmail.com wrote:
Hi, all
There
And in general, you can use:
str(NameOfObject)
To understand it's *str*ucture, and where the value resides.
Cheers.
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) |
On Tue, Aug 3, 2010 at 11:42 AM, Carlos Petti carlos.pe...@gmail.com wrote:
Dear list,
I have a list of matrices :
i1 - matrix(1:10, nrow = 2, ncol = 5)
i2 - matrix(11:20, nrow = 2, ncol = 5)
j - list(i1 = i1, i2 = i2)
I would like to attribute names to each dimension, for each matrix,
Hello,
I'm using the glmulti package to run models of all the possible combinations
of my variables. However, I am only interested in a few interactions between my
variables.
I have tried the equivalent of:
mod1-lm(y~a+b+c+a:b)
glmulti(mod1, level=1)
mod2-lm(y~a+b+c+a:b)
glmulti(mod2,
We are pleased to announce CRdata.org, a free open-source menu-driven
file-sharing and Cloud-Computing web-service for R users.
CRdata gives people without R expertise an intuitive graphical interface
to R, while R experts can use CRdata to share their algorithms with
non-experts, and to manage
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Nikhil Kaza
Sent: Tuesday, August 03, 2010 8:56 AM
To: Ron Michael
Cc: r-help@r-project.org
Subject: Re: [R] Need help on upper.tri()
try using Matrix package instead
Hello,
I'm using the glmulti package to run models of all the possible combinations
of my variables. However, I am only interested in a few interactions between my
variables.
I have tried the equivalent of:
mod1-lm(y~a+b+c+a:b)
glmulti(mod1, level=1)
mod2-lm(y~a+b+c+a:b)
glmulti(mod2,
There are other ways to make symmetric matrices:
1. mat + t(mat)
2. crossprod(mat)
3. tcrossprod(mat)
(1) is slightly faster than (2) and (3) (difference is trivial except for very
large matrices), but (2) and (3) are guranteed to give you a
positive-semidefinite (PSD) matrices, whereas
Hey guys and gals,
I searched through the forum and a bunch of R-mapping dedicated sites but
have not found what i know is quite elementary process, mapping more than
one layer on the same plot. I need to show some reference lines for the map
to make sense. I know the below wont work for
question again.
here is part of my original code for my research question. originally I used a
loop with matrix form to do the work and it works (though looks ugly). I have
tried the methods mentioned in the above posts and thought I did the same
thing.
but the new code does not work. the
I understand the question I am about to ask is rather vague and
depends on the task and my PC memory. However, I'll give it a try:
Let's assume the goal is just to read in the data frame into R and
then do some simple analyses with it (e.g., multiple regression of
some variables onto some - just
If you do an 'str' on your boject, you will probably find that TICKER
is a factor. This is probably not what you want. So when you create
the dataframe 'stock', use the parameter 'stringsAsFactors=FALSE' to
prevent the conversion to factors.
In the mean time, you can do:
stocks$TICKER -
Hi:
In addition to the previous replies, there is package R.Matlab and a
Matlab/R reference at CRAN by David Hiebeler under Contributed
Documentation.
HTH,
Dennis
On Mon, Aug 2, 2010 at 6:57 PM, leepama butch...@hanmail.net wrote:
I made some matlab codes...
Is there any method to perform
You probably don't want an object that is larger than about 25% of the
physical memory so that copies can be made during some processing. If
you are running on a 32-bit system which will limit you to at most 3GB
of memory, then your largest object should not be greater than 800MB.
If you want to
In the way you describe it, no.
A) There is no such thing as a zip folder. You are most likely being fooled
by the visual presentation of zip files in Windows Explorer, which
automatically simulates it as a folder, when it is really extracting bits of
data as needed to display it.
B) Because
Another solution
factor(port1[,2], labels=levels(stocks[,2]))
-
A R learner.
--
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Sent from the R help mailing list archive at Nabble.com.
On 03/08/2010 1:10 PM, Dimitri Liakhovitski wrote:
I understand the question I am about to ask is rather vague and
depends on the task and my PC memory. However, I'll give it a try:
Let's assume the goal is just to read in the data frame into R and
then do some simple analyses with it (e.g.,
Duncan Murdoch murdoch.dun...@gmail.com writes:
On 29/07/2010 6:18 PM, chipmaney wrote:
-Why does R recognize '[' as a function?
Because it is a function.
More explicitly, '[' is a string. sapply then calls match.fun to look
up that string to get the function named '['.
-Why does it
On 03.08.2010 17:48, Stephen Liu wrote:
Hi folks,
CPU - AMD X4 955
Onboard RAM - 8G
I'm prepared to learn R and am going to install R package on 64 bit Windows
running as VM (guest) on Oracle VituralBox. Please advise
1) I have 64 bit Win7 and Win Server 2008 R2 running as VM on
Thanks a lot, it's very helpful!
Dimitri
On Tue, Aug 3, 2010 at 1:53 PM, Duncan Murdoch murdoch.dun...@gmail.com wrote:
On 03/08/2010 1:10 PM, Dimitri Liakhovitski wrote:
I understand the question I am about to ask is rather vague and
depends on the task and my PC memory. However, I'll give
See also Omegahat package Rcompression (a copy of which for Windows is
on CRANextras).
But I would do this via unzip, modify, zip
On Tue, 3 Aug 2010, Jeff Newmiller wrote:
In the way you describe it, no.
A) There is no such thing as a zip folder. You are most likely
being fooled by the
And once one above the limit that Jim indicated - is there anything one can do?
Thank you!
Dimitri
On Tue, Aug 3, 2010 at 2:12 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Thanks a lot, it's very helpful!
Dimitri
On Tue, Aug 3, 2010 at 1:53 PM, Duncan Murdoch
On 03/08/2010 2:28 PM, Dimitri Liakhovitski wrote:
And once one above the limit that Jim indicated - is there anything one can do?
Yes, there are several packages for handling datasets that are too big
to fit in memory: biglm, ff, etc. You need to change your code to work
with them, so
Is this you want?
plot((1:20)^2,1:20,type=b,lwd=4,col=blue)
lines((1:20)^3,1:20,type=b,lwd=4,col=red)
-
A R learner.
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Sent from the R help mailing list archive at
I have just learned from the R Language Definiation
##Operators are special functions
# Examples
-(x,1:6)
[(x,3)
+(3,5)
'-'(3,6)
'/'(3,6)
'*'(3,6)
-
A R learner.
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Sent from the R help
I don't think one session can get information from another session, except
use load.
I simulate you situation within one session. Hope it helps.
#Simulate the first session
object1 - 1:10
save(object1,file=ob1)
rm(object1)
object1
#Simulate the second session
object2 - (1:10)*2
Hi:
On Tue, Aug 3, 2010 at 6:51 AM, haenl...@gmail.com wrote:
I'm sorry -- I think I chose a bad example. Let me start over again:
I want to estimate a moderated regression model of the following form:
y = a*x1 + b*x2 + c*x1*x2 + e
No intercept? What's your null model, then?
Based on
Dear Olga
An R session is conducted entirely in the RAM memory of your computer,
and each invocation of R will have its own memory space, not shared
with any other application, including another R session.
You will have to architect a scheme to allow one R session to find out
about events and
There are an infinite number of solutions for your example, so I hope you
really don't want to see all of them. In theory you could work up some code to
start showing them to you, but the sun will go nova and atomize you and your
computer before it shows all of them.
Expressing the infinite
you will have a better luck with R-sig-geo.
Unfortunately I could not find an easy way for polygon overlays.
plot(Tazshape)
lines(ugbshape) if ugbshape was a polyline instead of a polygon.
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
Hi all,
Is there a function that allow me to concatenate each value in a vector to a
string?
x-c(a,b,c,d)
output string = a+b+c+d
Thanks,
phoebe
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Try this:
paste(x, collapse = '+')
On Tue, Aug 3, 2010 at 5:34 PM, phoebe kong sityeek...@gmail.com wrote:
Hi all,
Is there a function that allow me to concatenate each value in a vector to
a
string?
x-c(a,b,c,d)
output string = a+b+c+d
Thanks,
phoebe
[[alternative HTML
Absolutely right.
But I think it's also worth adding that when the predictors _are_
correlated, the estimates of their coefficients depend on which are
included in the model. This means that one should generally not try to
interpret the individual coefficients, e.g. as a way to assess their
Thanks for all your comments!
@Dennis: Are there any thresholds that I can use to evaluate the Variance
Inflation Factor? I think I learned at some point that VIF should be less
than 10, but probably that is too conservative? You mentioned in your
example that a VIF of 13 is not big enough to
If I have a column with 2 levels, but one level has no remaining
observations. Can I remove the level?
Had intended to do it as listed below, but soon realized that even though
there are no observations, the level is still there.
For instance
summary(dbs3.train.sans.influential.obs$HAC)
biased regression coefficients is nonsense. The coefficients are
unbiased: their expectation (in the appropriate model) is the true
value of the parameters (when estimated by, e.g. least squares).
The problem is model selection. I suggest you consult a local
statistician, as you seem confused
On Wed, 04 Aug 2010, Erin Hodgess wrote:
Thank you!
Now, here is the code for the FORTRAN subroutine:
subroutine cov1(n1,na,x,z,yy,ac)
integer n1,na
real x(n1),ac(na+1),z,yy
do 5 j=1,(na+1)
do 10 i=1,(n1-j)
if(j.eq.1)ac(j)=yy
GL wrote:
If I have a column with 2 levels, but one level has no remaining
observations. Can I remove the level?
What is a 'column'? An element of a data.frame?
Does the following help?
f1 - factor(L1, levels = c(L1, L2))
levels(f1)
f1 - factor(f1)
levels(f1)
In absence of a
Dear R Help,
Is there a way to set constraints on the fixed effects parameters?
Thank you.
Sincerely,
Sarah Marston
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How can I relevel a liner model with a numeric by dummy variable interaction
to extract the model estimate, std. error, and t-value for the reference
factor?
Thank you,
Christy
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Actually, you probably want to remove the remaining level -- that is,
remove the variable altogether, since if it has only a single value
its effect is indistinguishable from the overall mean.
Again, complying with the posting guide would be advisable.
Bert Gunter
Genentech Nonclinical
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